Let \(G\) be a group and \(N\) a normal subgroup of \(G\). Then we may define the quotient group \(G/N\) to be the set of left cosets \(gN\) of \(N\) in \(G\), with the group operation that \(gN + hN = (gh)N\). This is well-defined if and only if \(N\) is normal.

Proof

\(N\) normal implies \(G/N\) well-defined

Recall that \(G/N\) is well-defined if “it doesn’t matter which way we represent a coset”: whichever coset representatives we use, we get the same answer.

Suppose \(N\) is a normal subgroup of \(G\). We need to show that given two representatives \(g_1 N = g_2 N\) of a coset, and given representatives \(h_1 N = h_2 N\) of another coset, that \((g_1 h_1) N = (g_2 h_2)N\).

So given an element of \(g_1 h_1 N\), we need to show it is in \(g_2 h_2 N\), and vice versa.

Let \(g_1 h_1 n \in g_1 h_1 N\); we need to show that \(h_2^{-1} g_2^{-1} g_1 h_1 n \in N\), or equivalently that \(h_2^{-1} g_2^{-1} g_1 h_1 \in N\).

But \(g_2^{-1} g_1 \in N\) because \(g_1 N = g_2 N\); let \(g_2^{-1} g_1 = m\). Similarly \(h_2^{-1} h_1 \in N\) because \(h_1 N = h_2 N\); let \(h_2^{-1} h_1 = p\).

Then we need to show that \(h_2^{-1} m h_1 \in N\), or equivalently that \(p h_1^{-1} m h_1 \in N\).

Since \(N\) is closed under conjugation and \(m \in N\), we must have that \(h_1^{-1} m h_1 \in N\); and since \(p \in N\) and \(N\) is closed under multiplication, we must have \(p h_1^{-1} m h_1 \in N\) as required.

\(G/N\) well-defined implies \(N\) normal

Fix \(h \in G\), and consider \(hnh^{-1} N + hN\). Since the quotient is well-defined, this is \((hnh^{-1}h) N\), which is \(hnN\) or \(hN\) (since \(nN = N\), because \(N\) is a subgroup of \(G\) and hence is closed under the group operation). But that means \(hnh^{-1}N\) is the identity element of the quotient group, since when we added it to \(hN\) we obtained \(hN\) itself.

That is, \(hnh^{-1}N = N\). Therefore \(hnh^{-1} \in N\).

Since this reasoning works for any \(h \in G\), it follows that \(N\) is closed under conjugation by elements of \(G\), and hence is normal.